$\begin{aligned} &g(x)=\dfrac{1}{3}(x+1) \\\\ &h(x)=\dfrac{x^2-1}{x+3} \end{aligned}$ $(g\circ h) (5)=$
Explanation: Let's start by rewriting $(g\circ h) (5)$ as $g(h(5))$. When evaluating composite functions, we work our way inside out. To evaluate $g(h(5))$, let's first evaluate $h(5)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $h({5})$. $\begin{aligned}h(x)&=\dfrac{x^2-1}{x+3}\\\\ h({5})&=\dfrac{({5})^2-1}{({5})+3} ~~~~~~~~~~\text{Plug in }x={5}\\\\ &=\dfrac{24}{8}\\\\ &={3}\end{aligned}$ We now know that $g(h({5}))$ is the same as $g(3)$ because $h({5}) = 3$. Let's evaluate $g({3})$. $\begin{aligned}g(x)&=\dfrac{1}{3}(x+1)\\\\ g({{3}})&=\dfrac{1}{3}(({3})+1)~~~~~~~~~~\text{Plug in }x={3}\\\\ &=\dfrac{1}{3}(4)\\\\ &=\dfrac{4}{3}\end{aligned}$ The answer: $(g\circ h)(5) =\dfrac{4}{3}$